157-7-3化學反應機理和催化劑

1、第第 4 節(jié)反應機理節(jié)反應機理所謂基元反應是指反應物分子一步直接轉化為產(chǎn)物的反應。 如：NO2 CO NO CO2 反應物NO2 分子和CO分子經(jīng)過一次碰撞就轉變成為產(chǎn)物NO分子和CO2 。 基元反應是動力學研究中的最簡單的反應，反應過程中沒有任何中間產(chǎn)物。一、基本概念Elementary reactions are steps of molecular events showing how reactions proceed. This type of description is a mechanism. The mechanism for the reaction between CO

2、and NO2 is proposed to beStep 1 NO2 + NO2 NO3 + NO(an elementary reaction)Step 2 NO3 + CO NO2 + CO2(an elementary reaction)Add these two equations led to the overall reactionNO2 + CO = NO + CO2(overall reaction)A mechanism is a proposal to explain the rate law, and it has to satisfy the rate law. A

3、satisfactory explanation is not a proof.例如： H2 ( g ) I2 ( g ) 2 HI ( g )實驗上或理論上都證明，它并不是一步完成的基元反應，它的反應歷程可能是如下兩步基元反應： I2 I I （快） H2 2 I 2 HI （慢） 化學反應的速率由反應速率慢的基元反應決定。 基元反應或復雜反應的基元步驟中發(fā)生反應所需要的微粒（分子、原子、離子）的數(shù)目一般稱為反應的分子數(shù)。分子數(shù)Molecularity of Elementary ReactionsThe total order of rate law in an elementary re

4、action is molecularity.The rate law of elementary reaction is derived from the equation. The order is the number of reacting molecules because they must collide to react.A molecule decomposes by itself is a unimolecular reaction (step);two molecules collide and react is a bimolecular reaction (step)

5、; &three molecules collide and react is a termolecular reaction (step).O3 O2 + Orate = k O3NO2 + NO2 NO3 + NOrate = k NO22Br + Br + Ar Br2 + Ar*rate = k Br2ArCaution: Derive rate laws this way only for elementary reactions.單分子反應SO2Cl2 的分解反應 SO2Cl2 SO2 Cl2雙分子反應NO2 的分解反應 2 NO2 2 NO O2三分子反應 HI 的生成反

6、應 H2 2 I 2 HI 四分子或更多分子碰撞而發(fā)生的反應尚未發(fā)現(xiàn)。Elementary Reactions are Molecular EventsN2O5 NO2 + NO3 NO + O2 + NO2 NO2 + NO3A mechanism is a collection of elementary steps devise to explain the the reaction in view of the observed rate law. You need the skill to derive a rate law from a mechanism, but proposi

7、ng a mechanism is task after you have learned more chemistry For the reaction, 2 NO2 (g) + F2 (g) 2 NO2F (g), the rate law is,rate = k NO2 F2 .Can the elementary reaction be the same as the overall reaction?If they were the same the rate law would have been rate = k NO22 F2,Therefore, they the overa

8、ll reaction is not an elementary reaction. Its mechanism is proposed next.The rate determining step is the slowest elementary step in a mechanism, and the rate law for this step is the rate law for the overall reaction.The (determined) rate law is,rate = k NO2 F2,for the reaction, 2 NO2 (g) + F2 (g)

9、 2 NO2F (g), and a two-step mechanism is proposed:i NO2 (g) + F2 (g) NO2F (g) + F (g)ii NO2 (g) + F (g) NO2F (g)Which is the rate determining step?Answer:The rate for step i is rate = k NO2 F2, which is the rate law, this suggests that step i is the rate-determining or the s-l-o-w step.反應機理中的慢反應步驟決定

10、總反應的速率！反應機理中的慢反應步驟決定總反應的速率！二、如何由給出的反應機理推導出速率方程例1、The decomposition of H2O2 in the presence of I follow this mechanism,iH2O2 + I k1 H2O + IO slow ii H2O2 + IO k2 H2O + O2 + I fastWhat is the rate law?SolutionThe slow step determines the rate, and the rate law is:rate = k1 H2O2 I Since both H2O2 and I

11、 are measurable in the system, this is the rate law.例例2、Derive the rate law for the reaction, H2 + Br2 = 2 HBr, from the proposed mechanism:i Br2 2 Brfast equilibrium (k1, k-1)iiH2 + Br k2 HBr + H slow iii H + Br k3 HBr fastSolution:The fast equilibrium condition simply says thatk1 Br2 = k-1 Br2andB

12、r = (k1/k-1 Br2)The slow step determines the rate law,rate = k2 H2 Br Br is an intermediate = k2 H2 (k1/k-1 Br2) = k H2 Br2 ; k = k2 (k1/k-1) M- s -1total order 1.5explain快速平衡假設法！例3、The decomposition of N2O5 follows the mechanism:1N2O5 NO2 + NO3fast equilibrium 2NO2 + NO3 k2 NO + O2 + NO2slow3NO3 +

13、NO k3 NO2 + NO2fastDerive the rate law.Solution:The slow step determines the rate,rate = k2 NO2 NO3 NO2 & NO3 are intermediateFrom 1, we have NO2 NO3 = KK, equilibrium constant N2O5 K differ from kThus, rate = K k2 N2O5穩(wěn)態(tài)近似法！以假設中間產(chǎn)物的濃度恒定不變?yōu)榛A！即、中間產(chǎn)物的生成速率與其消耗速率相等。Rate of producing the intermediat

14、e, Rprod, is the same as its rate of consumption, Rcons.Rprod = RconsIntermediatetimeRprod RconsBe able to apply the steady-state approximation to derive rate laws假設 H2 + I2 2 HI的反應機理如下：Step (1)I2 k1 2 IStep (1)2 I k-1 I2Step (2) H2 + 2 I k2 2 HIDerive the rate law.Derivation:rate = k2 H2 I 2 (cause

15、 this step gives products！)but I is an intermediate, this is not a rate law yet.Since k1 I2(= rate of producing I) = k-1 I2 + k2 H2 I2(= rate of consuming I)Thus, k1 I2 I2 = k-1 + k2 H2 rate = k1 k2 H2 I2 / k-1 + k2 H2 Steady stateFrom the previous result:k1 k2 H2 I2rate = k-1 + k2 H2 Discussion:(i) If k-1 k2 H2 then k-1 + k2 H2 = k2 H2 ,then rate = k1 k2 H2 I2 / k2